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jaeseo222
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Apr 4, 2022
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| for (int i = 1; i < n; i++) { | ||
| int m = 0; | ||
| for (int j = 0; j < i; j++) { //i번 앞 원소들 순회 | ||
| if (a[j] < a[i]) { //증가하는 수열이 되면 | ||
| m = max(m, dp[j]); //길이 최댓값 업데이트 | ||
| } | ||
| } | ||
| dp[i] = m + 1; //구한 길이에 해당 원소 포함되니 + 1 | ||
| } | ||
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| // for(int i: dp){ | ||
| // cout<<i<<' '; | ||
| // } | ||
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| //출력 | ||
| cout << *max_element(dp.begin(), dp.end()); |
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p2.
증가하는 수열인 것을 확인하고 m을 갱신해주시며 dp에 저장해주셨군요! 좋아요!!🤗🤗
그런데, 증가하는 수열인 것을 확인했다면 m에 저장해놓는 것보다 바로 dp를 갱신해주는 건 어떨까요?
dp의 포인트는 각 인덱스마다의 값을 최댓값으로 (혹은 문제에 따라 최솟값) 갱신해주는 것입니다!
따라서, dp의 맨 마지막 인덱스엔 구하려는 최댓값이 자연스럽게 남아있게 되는 거죠! max_element으로 가장 큰 답을 구해주지 않아도 바로 참조가능하죠.🤗🤗
강의자료를 참고하시면 더 이해가 잘 되실 것 같습니다!
| dp[1] = 1; | ||
| dp[2] = 3; //dp[3] = 3 + (1+1) = 5 | ||
| for (int i = 3; i <= n; i++) { | ||
| dp[i] = (dp[i - 2] * 2 + dp[i - 1]) % 10007; |
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내용 & 질문
<기존 제출>